Solve for $t$, $ -\dfrac{2t}{3t - 9} = \dfrac{1}{t - 3} - \dfrac{4}{t - 3} $
Explanation: First we need to find a common denominator for all the expressions. This means finding the least common multiple of $3t - 9$ $t - 3$ and $t - 3$ The common denominator is $3t - 9$ The denominator of the first term is already $3t - 9$ , so we don't need to change it. To get $3t - 9$ in the denominator of the second term, multiply it by $\frac{3}{3}$ $ \dfrac{1}{t - 3} \times \dfrac{3}{3} = \dfrac{3}{3t - 9} $ To get $3t - 9$ in the denominator of the third term, multiply it by $\frac{3}{3}$ $ -\dfrac{4}{t - 3} \times \dfrac{3}{3} = -\dfrac{12}{3t - 9} $ This give us: $ -\dfrac{2t}{3t - 9} = \dfrac{3}{3t - 9} - \dfrac{12}{3t - 9} $ If we multiply both sides of the equation by $3t - 9$ , we get: $ -2t = 3 - 12$ $ -2t = -9$ $ -2t = -9 $ $ t = \dfrac{9}{2}$